Overview
This chapter introduces inverse trigonometric functions (principal values of arcsin, arccos, arctan, arccot, arcsec, arccosec) and their role as the inverses of standard trigonometric functions after restricting domains to make them one-to-one. It explains domains and principal-value ranges, basic algebraic relations among inverse trig functions, their graphs, and how to manipulate and simplify expressions involving them. The chapter is important because inverse trig functions frequently appear in calculus (differentiation and integration), in solving equations and inequalities, and in modelling problems in physics and engineering. Students will learn definitions and principal branches, evaluate and simplify inverse-trig expressions, prove identities, compute derivatives and integrals involving inverse trig functions, solve equations and inequalities, and use composition relations (e.g., sin(sin^{-1} x) versus sin^{-1}(sin x)) while carefully handling domain restrictions. Emphasis is placed on monotonicity, continuity, standard formulas (addition formulas for arctan), and exam-focused problem techniques (domain checks, sign considerations, and algebraic substitutions).
Learning Objectives
- Define principal values and principal-value branches of inverse trigonometric functions (arcsin, arccos, arctan, arccot, arcsec, arccosec).
- State the domain and range of arcsin x, arccos x, arctan x, arccot x, arcsec x and arccosec x and justify them.
- Derive and verify standard relations such as arcsin x + arccos x = π/2 and conditions for the formula arctan x + arctan y = arctan((x+y)/(1−xy)).
- Explain the behaviour of compositions like sin(arcsin x) and arcsin(sin x), indicating when the identity holds and when principal-value adjustment is needed.
- Evaluate exact values of inverse trigonometric expressions for standard angles and simple algebraic numbers.
- Simplify algebraic expressions involving inverse trigonometric functions using identities and domain/range considerations.
- Solve equations involving inverse trigonometric functions (linear and simple nonlinear) within given intervals and provide all admissible solutions.
- Differentiate inverse trigonometric functions and find derivatives of composite functions using the chain rule (e.g., d/dx[arcsin u(x)], d/dx[arctan u(x)]).
Topics in this chapter
14 topics · tap a topic title to jump straight to it.
Basic definitions and principal branches
Basic definitions and principal branches
Core Principle: arcsin: domain [−1,1], range [−π/2, π/2], sin(arcsin x) = x for x ∈ [−1,1], arcsin(sin y) = y for y ∈ [−π/2, π/2].
What is an inverse trigonometric function? A trigonometric function (sin, cos, tan, etc.) is not one-to-one on its whole domain, so it does not have a global inverse. To define an inverse, we restrict the domain of the original function to a principal branch on which it is one-to-one and then define the inverse on that restricted range. The inverse trigonometric function returns the principal angle whose trig value equals a given number.
Why principal branches? Without restricting the domain, an equation like sin y = x has infinitely many solutions y = arcsin x + 2kπ and y = (π - arcsin x) + 2kπ. A principal branch picks one representative (principal value) from these infinitely many values so the inverse becomes a single-valued function.
Standard principal branches (domains and ranges):
- y = arcsin x: domain x ∈ [−1, 1], range y ∈ [−π/2, π/2].
- y = arccos x: domain x ∈ [−1, 1], range y ∈ [0, π].
- y = arctan x: domain x ∈ (−∞, ∞), range y ∈ (−π/2, π/2).
- y = arccot x: domain x ∈ (−∞, ∞), range y ∈ (0, π).
- y = arcsec x: domain x ∈ (−∞, −1] ∪ [1, ∞), range y ∈ [0, π], y ≠ π/2.
- y = arccsc x: domain x ∈ (−∞, −1] ∪ [1, ∞), range y ∈ [−π/2, π/2], y ≠ 0.
How to obtain a graph of an inverse function: Graph of an inverse function is the reflection of the graph of the original function (restricted to the principal branch) about the line y = x. So: restrict the trig function to its principal domain (e.g., sin on [−π/2, π/2], cos on [0, π], tan on (−π/2, π/2]), sketch it, then reflect about y = x to get graph of arcsin, arccos, arctan respectively.
Key properties (one-line summary): For x in the domain of the inverse, trig(inverse(x)) = x. For y in the principal range, inverse(trig(y)) = y. Principal inverses are monotone on their domains (arcsin & arctan increasing, arccos decreasing).
- If sin θ = 0.6, the principal value θ = arcsin(0.6) ≈ 0.6435 rad ≈ 36.87° (since arcsin range is [−π/2, π/2]).
- If tan θ = 2, the principal value θ = arctan(2) ≈ 1.1071 rad ≈ 63.434° (arctan range (−π/2, π/2)).
- If sec θ = −2, then θ = arcsec(−2) = arccos(1/(−2)) = arccos(−1/2) = 2π/3 ≈ 2.0944 rad (arcsec principal range [0, π], ≠ π/2).
- Check identity: arcsin(1/2) = π/6 and arccos(1/2) = π/3; arcsin(1/2) + arccos(1/2) = π/2.
- \[arcsin: domain [−1,1]\]\[range [−π/2, π/2]\]\[sin(arcsin x) = x for x ∈ [−1,1]\]\[arcsin(sin y) = y for y ∈ [−π/2, π/2].\]
- \[arccos: domain [−1,1]\]\[range [0, π]\]\[cos(arccos x) = x for x ∈ [−1,1]\]\[arccos(cos y) = y for y ∈ [0, π].\]
- \[arctan: domain (−∞, ∞)\]\[range (−π/2, π/2)\]\[tan(arctan x) = x\]\[arctan(tan y) = y for y ∈ (−π/2, π/2).\]
- \[arccot: domain (−∞, ∞)\]\[range (0, π)\]\[cot(arccot x) = x.\]
- \[arcsec: domain |x| ≥ 1\]\[range [0, π] \ {π/2}\]\[sec(arcsec x) = x\]\[arcsec x = arccos(1/x) for |x| ≥ 1.\]
- \[arccsc: domain |x| ≥ 1\]\[range [−π/2, π/2] \ {0}\]\[csc(arccsc x) = x\]\[arccsc x = arcsin(1/x) for |x| ≥ 1.\]
Inverse function concept and need for restriction
Inverse function concept and need for restriction
Core Principle: Range and principal branches: arcsin: domain [-1,1], range [-π/2,π/2]; arccos: domain [-1,1], range [0,π]; arctan: domain (-∞,∞), range (-π/2,π/2).
What is an inverse function? For a function f, an inverse f^{-1} undoes the action of f: f^{-1}(f(x)) = x for every x in the domain of f, and f(f^{-1}(y)) = y for every y in the range of f. An inverse exists only if f is one-to-one (injective) and onto its range.
Why restriction is needed for trigonometric functions? Standard trigonometric functions (sin, cos, tan) are periodic and not one-to-one on their full domains. For example, sin x = sin(x + 2π) and many different x give the same sine value. Because of this, the inverse (e.g., arcsin) is not well-defined unless we restrict the domain of the original function to a suitable interval where it is one-to-one and covers all possible values (its range).
Principal value branches (standard restrictions):
- sin x is restricted to x ∈ [-π/2, π/2] to define y = arcsin x with range y ∈ [-π/2, π/2].
- cos x is restricted to x ∈ [0, π] to define y = arccos x with range y ∈ [0, π].
- tan x is restricted to x ∈ (-π/2, π/2) to define y = arctan x with range y ∈ (-π/2, π/2).
These chosen intervals are called principal branches; they make each function monotonic (strictly increasing or decreasing) and satisfy the horizontal-line test, so an inverse exists and is unique for each y in the function's range.
Important caution: While sin(arcsin x) = x for x in [-1,1], the composition arcsin(sin x) is not always equal to x for all x — it gives the principal value in [-π/2, π/2] that has the same sine as x. The same applies to arccos(cos x) and arctan(tan x).
- Measuring angle from sensor: A tilt sensor returns sin(θ) = 0.6. To find the tilt angle θ within the principal range, use θ = arcsin(0.6) ≈ 0.6435 rad (≈ 36.87°).
- Navigation/bearing: If a bearing calculation yields tan(θ) = 1, then θ = arctan(1) = π/4 (45°) inside (-π/2, π/2). If you are using angles outside that interval, you must adjust by adding π or using quadrant information.
- Triangle problem: Given opposite/hypotenuse = 3/5, θ = arcsin(3/5). This gives the acute angle of the triangle directly because arcsin returns values in [-π/2, π/2].
- Computer graphics: To recover an angle from the y-component of a rotated unit vector (y = sin θ), use arcsin(y) but restrict to the principal branch or use sign of x-component to determine the correct quadrant.
- \[Range and principal branches: arcsin: domain [-1,1]\]\[range [-&pi\]\[/2,&pi\]\[/2]\]\[arccos: domain [-1,1]\]\[range [0,&pi\]\[]\]\[arctan: domain (-&infin\]\[,&infin\]\[)\]\[range (-&pi\]\[/2,&pi\]\[/2).\]
- \[Composition identities (principal-value aware): sin(arcsin x) = x (x &in\]\[[-1,1])\]\[cos(arccos x) = x (x &in\]\[[-1,1])\]\[tan(arctan x) = x (x &in\]\[R).\]
- \[Inverse-to-original (may differ by multiples of &pi\]\[if outside principal range): arcsin(sin x) = x' where x' is the principal value in [-&pi\]\[/2,&pi\]\[/2] with the same sine as x.\]
- \[Derivatives (useful in calculus): d/dx[arcsin x] = 1/√(1 - x^2) for x in (-1,1)\]\[d/dx[arccos x] = -1/√(1 - x^2)\]\[d/dx[arctan x] = 1/(1 + x^2).\]
Domain and range
Domain and range
Core Principle: Domain and range (principal): arcsin x: domain [−1,1], range [−π/2, π/2] arccos x: domain [−1,1], range [0, π] arctan x: domain R, range (−π/2, π/2) arccot x: domain R, range (0, π) arcsec x: domain (−∞,−1] ∪ [1,∞), range [0,π] \ {π/2} arccosec x: domain (−∞,−1] ∪ [1,∞), range [−π/2,π/2] \ {0}
Meaning: For a function y = f(x), the domain is the set of all x-values for which f(x) is defined; the range is the set of all possible y-values (outputs). For inverse trigonometric functions, we define principal branches (restrict domains of the original trig functions) so that each inverse is single-valued.
Why restriction is needed: Standard trigonometric functions are not one-to-one on their natural domains (they are periodic). To define an inverse function f^{-1}, we restrict the original function to an interval where it is one-to-one and onto a convenient range. Those standard principal restrictions give the usual domains and ranges of arcsin, arccos, arctan, arccot, arcsec and arccosec used in Class 12.
Principal branches (summary): We choose intervals for sine, cosine, tangent etc. so that they are monotonic and cover all possible outputs once. Then we define the inverse functions on the corresponding output sets. The common (CBSE) choices are:
- y = sin^{-1} x (arcsin x): domain x \in [-1,1], range y \in [-\pi/2,\pi/2]
- y = cos^{-1} x (arccos x): domain x \in [-1,1], range y \in [0,\pi]
- y = tan^{-1} x (arctan x): domain x \in (-\infty,\infty), range y \in (-\pi/2,\pi/2)
- y = cot^{-1} x (arccot x): domain x \in (-\infty,\infty), range y \in (0,\pi)
- y = sec^{-1} x (arcsec x): domain x \in (-\infty,-1]\cup [1,\infty), range y \in [0,\pi]\setminus\{\pi/2\}
- y = cosec^{-1} x (arccosec x): domain x \in (-\infty,-1]\cup [1,\infty), range y \in [-\pi/2,\pi/2]\setminus\{0\}
Important remarks: For any inverse trig function f^{-1} and its original f, f(f^{-1}(x)) = x for x in the domain of f^{-1}. But f^{-1}(f(x)) = x only when x lies in the restricted principal interval (the chosen range) of f.
- arcsin(1/2) = π/6 because sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2].
- arccos(−1) = π because cos(π) = −1 and π ∈ [0, π].
- arctan(√3) = π/3 because tan(π/3) = √3 and π/3 ∈ (−π/2, π/2).
- arcsec(2) = π/3 because sec(y)=2 ⇒ cos(y)=1/2 and principal cos–value is y=π/3 ∈ [0,π]\{π/2}.
- arccosec(−2) = −π/6 because csc(y)=−2 ⇒ sin(y)=−1/2 and principal sin–value is y=−π/6 ∈ [−π/2,π/2]\{0}.
- \[Domain and range (principal): arcsin x: domain [−1,1]\]\[range [−π/2, π/2] arccos x: domain [−1,1]\]\[range [0, π] arctan x: domain R\]\[range (−π/2, π/2) arccot x: domain R\]\[range (0, π) arcsec x: domain (−∞,−1] ∪ [1,∞)\]\[range [0,π] \ {π/2} arccosec x: domain (−∞,−1] ∪ [1,∞)\]\[range [−π/2,π/2] \ {0}\]
- \[Basic inversion identities (with domain/range conditions): sin(arcsin x) = x for x ∈ [−1,1]\]\[arcsin(sin θ) = θ only if θ ∈ [−π/2,π/2]\]\[Similar remarks hold for cos\]\[tan\]\[etc.\]\[with their principal ranges.\]
- \[Common useful relations: arcsin x + arccos x = π/2 for x ∈ [−1,1]. arctan x + arctan(1/x) = π/2 for x>0 (care with signs when x<0). arctan x + arccot x = π/2 for all real x (using principal ranges).\]
- \[Derivatives (often used in Class 12): d/dx[arcsin x] = 1/√(1−x^2), |x|<1\]\[d/dx[arccos x] = −1/√(1−x^2), |x|<1\]\[d/dx[arctan x] = 1/(1+x^2)\]\[x∈R.\]
Principal values and notations
Principal values and notations
Core Principle: Principal ranges: arcsin x ∈ [−π/2, π/2], arccos x ∈ [0, π], arctan x ∈ (−π/2, π/2), arccot x ∈ (0, π), arcsec x ∈ [0, π] \ {π/2}, arccsc x ∈ [−π/2, π/2] \ {0}.
What is the principal value?
Inverse trigonometric functions (like sin^{-1}, cos^{-1}, tan^{-1}) are multi-valued because trigonometric functions are periodic. To get a single-valued (well-defined) function we choose one specific range (called the principal branch or principal value range). The value of the inverse function taken from that chosen range is called the principal value.
Why we need principal values
If sin θ = x there are infinitely many θ values (because of periodicity). For a function denoted by arcsin x (or sin^{-1} x) we pick one of those θ values — the one in the standard principal range — so arcsin becomes a proper function.
Standard notations
Common notations for inverse trig functions:
- arcsin x or sin^{-1} x (read "arc sine" or "inverse sine")
- arccos x or cos^{-1} x
- arctan x or tan^{-1} x
- arccot x or cot^{-1} x
- arcsec x or sec^{-1} x
- arccsc x or csc^{-1} x
Note: sin^{-1} x is commonly used for arcsin x in school texts but might be confused with 1/sin x (cosecant) in other contexts. Prefer writing arcsin, arccos, arctan in careful work.
Principal value ranges (standard CBSE / NCERT choices)
- y = arcsin x : domain x ∈ [−1, 1], principal value range y ∈ [−π/2, π/2]
- y = arccos x : domain x ∈ [−1, 1], principal value range y ∈ [0, π]
- y = arctan x : domain x ∈ (−∞, ∞), principal value range y ∈ (−π/2, π/2)
- y = arccot x : domain x ∈ (−∞, ∞), principal value range y ∈ (0, π)
- y = arcsec x : domain |x| ≥ 1, principal value range y ∈ [0, π], y ≠ π/2
- y = arccsc x : domain |x| ≥ 1, principal value range y ∈ [−π/2, π/2], y ≠ 0
Principal value vs general solution
If α is the principal value (e.g. α = arcsin x), the full set of angles θ satisfying the original trig equation can be written using α plus integer multiples of periods. Examples:
- sin θ = x: θ = α + 2kπ or θ = π − α + 2kπ, where α = arcsin x ∈ [−π/2, π/2]
- cos θ = x: θ = ±β + 2kπ, where β = arccos x ∈ [0, π]
- tan θ = x: θ = γ + kπ, where γ = arctan x ∈ (−π/2, π/2)
Important remarks
Using the principal value ranges ensures that compositions like sin(arcsin x) = x (for x in domain) and arcsin(sin y) = y only hold when y is restricted to the principal range. Always check the principal-range condition when you invert or compose trig functions.
- Angle of elevation: If a ramp rises 1 m over 4 m horizontal, slope = 1/4, angle θ = arctan(1/4) (θ ∈ (−π/2, π/2)).
- Survey measurement: If a measured sine value of an angle is 0.5, principal value is arcsin(0.5) = π/6. General solutions are π/6 + 2kπ and 5π/6 + 2kπ.
- Electronics (phase angle): For an R–C circuit tan φ = X_C / R; φ = arctan(X_C/R) taken in (−π/2, π/2) gives the principal phase angle.
- \[Principal ranges: arcsin x ∈ [−π/2, π/2]\]\[arccos x ∈ [0, π]\]\[arctan x ∈ (−π/2, π/2)\]\[arccot x ∈ (0, π)\]\[arcsec x ∈ [0, π] \ {π/2}\]\[arccsc x ∈ [−π/2, π/2] \ {0}.\]
- \[Inverse–direct identities (within domains): sin(arcsin x) = x (|x| ≤ 1)\]\[cos(arccos x) = x (|x| ≤ 1)\]\[tan(arctan x) = x (all x).\]
- \[Composition with principal-range restriction: arcsin(sin θ) = θ if θ ∈ [−π/2, π/2]\]\[arccos(cos θ) = θ if θ ∈ [0, π]\]\[arctan(tan θ) = θ if θ ∈ (−π/2, π/2).\]
- \[Relations: arcsin x + arccos x = π/2 for x ∈ [−1,1]\]\[arctan x + arccot x = π/2 for real x.\]
- \[Reciprocal relations (with domain care): arcsec x = arccos(1/x) (|x| ≥ 1) and arccsc x = arcsin(1/x) (|x| ≥ 1).\]
- \[General solutions: if α = arcsin x\]\[then sin θ = x ⇒ θ = α + 2kπ or θ = π − α + 2kπ\]\[if γ = arctan x\]\[tan θ = x ⇒ θ = γ + kπ (k ∈ ℤ).\]
Graphs of inverse trigonometric functions
Graphs of inverse trigonometric functions
Core Principle: Basic compositions: sin(arcsin x) = x (for x in [-1,1]); cos(arccos x) = x (for x in [-1,1]); tan(arctan x) = x (for all real x).
Inverse trigonometric functions (also called arc-functions) give the angle whose trigonometric ratio equals a given number. Graphs of inverse trigonometric functions are obtained by reflecting the graphs of the corresponding restricted trigonometric functions about the line y = x, because if y = f^{-1}(x) then x = f(y).
Principal branches (used in Class 12 / CBSE) and key qualitative features:
- y = arcsin x (sin^{-1}x): domain = [-1, 1], range = [-π/2, π/2]. Graph is increasing, continuous on the domain, odd (arcsin(-x) = -arcsin x). Key points: arcsin(-1) = -π/2, arcsin(0)=0, arcsin(1)=π/2.
- y = arccos x (cos^{-1}x): domain = [-1, 1], range = [0, π]. Graph is decreasing, continuous. Key points: arccos(-1)=π, arccos(0)=π/2, arccos(1)=0. Relation to arcsin: arccos x = π/2 - arcsin x.
- y = arctan x (tan^{-1}x): domain = (-∞, ∞), range = (-π/2, π/2). Graph is increasing, continuous, with horizontal asymptotes y = ±π/2. Key points: arctan(0)=0, arctan(1)=π/4, arctan(-1)=-π/4.
- y = arccot x (cot^{-1}x): commonly taken with range (0, π). Graph is decreasing, horizontal asymptotes y=0 and y=π (principal branch). Relationship: arccot x = π/2 - arctan x (depending on branch conventions).
- y = arcsec x, y = arccsc x: defined for |x| ≥ 1. arcsec range often taken as [0,π] \ {π/2}, arccsc range as [-π/2,π/2] \ {0} (branch conventions vary). Their graphs have vertical exclusions near x in (-1,1) and behave like reflections of restricted sec and csc.
How to obtain each graph: restrict the original trig function to its principal interval where it is one-to-one, sketch that restricted graph, then reflect about y = x. For example: restrict y = sin x to [-π/2,π/2], reflect to get y = arcsin x.
- 1) Angle from a given sine: If sin θ = 0.6 and θ is principal, θ = arcsin(0.6) ≈ 0.6435 rad ≈ 36.87°. The graph of y = arcsin x shows this angle as the y-value at x = 0.6.
- 2) Angle of elevation (slope): A ramp has slope m = 0.75. The angle of inclination θ = arctan(0.75) ≈ 0.6435 rad ≈ 36.87°. The arctan graph gives the angle for any real slope and shows horizontal asymptotes as slope → ±∞.
- 3) Phase angle in AC circuit: For an R-L circuit, tan φ = X_L / R = 3/4. The phase angle φ = arctan(3/4) ≈ 36.87°. Use the arctan graph to interpret how φ changes as the reactance/resistance ratio varies.
- \[Basic compositions: sin(arcsin x) = x (for x in [-1,1])\]\[cos(arccos x) = x (for x in [-1,1])\]\[tan(arctan x) = x (for all real x).\]
- \[Inverse compositions: arcsin(sin y) = y for y in [-π/2, π/2]\]\[arccos(cos y) = y for y in [0, π]\]\[arctan(tan y) = y for y in (-π/2, π/2).\]
- \[Relations: arccos x = π/2 − arcsin x\]\[arctan x + arccot x = π/2 (with principal-value conventions).\]
- \[Derivatives (useful in calculus): d/dx[arcsin x] = 1/√(1−x^2) for |x|<1\]\[d/dx[arccos x] = −1/√(1−x^2)\]\[d/dx[arctan x] = 1/(1+x^2).\]
- \[Even/odd: arcsin x and arctan x are odd: f(−x) = −f(x). arccos x is neither odd nor even (but arccos(−x) = π − arccos x).\]
Basic identities and relations
Basic identities and relations
Core Principle: Principal ranges: sin⁻¹ x ∈ [−π/2, π/2], cos⁻¹ x ∈ [0, π], tan⁻¹ x ∈ (−π/2, π/2), cot⁻¹ x ∈ (0, π).
What the topic covers
This topic gives the fundamental properties, principal–value ranges and algebraic relations among inverse trigonometric functions (arcsin, arccos, arctan, arccot, arcsec, arccosec). These identities are used to simplify expressions containing inverse trig functions and to convert between different inverse functions.
Principal ranges (CBSE standard)
sin⁻¹ x ∈ [−π/2, π/2], cos⁻¹ x ∈ [0, π], tan⁻¹ x ∈ (−π/2, π/2), cot⁻¹ x ∈ (0, π), sec⁻¹ x ∈ [0, π] \ {π/2}, csc⁻¹ x ∈ [−π/2, π/2] \ {0}.
Basic composition rules
For values in the domain where the inverse is defined:
sin(sin⁻¹ x) = x (for −1 ≤ x ≤ 1) , cos(cos⁻¹ x) = x, tan(tan⁻¹ x) = x (for all real x).
Inverse of a function returns the original argument only if that argument lies in the principal range. For example: sin⁻¹(sin θ) = θ only when θ ∈ [−π/2, π/2]; otherwise the value is the principal value equivalent.
Key algebraic relations (fundamental)
sin⁻¹ x + cos⁻¹ x = π/2 (for −1 ≤ x ≤ 1)
tan⁻¹ x + cot⁻¹ x = π/2 (for all real x, with principal ranges)
sec⁻¹ x + csc⁻¹ x = π/2 (for |x| ≥ 1, with principal ranges)
Parity / symmetry
sin⁻¹(−x) = − sin⁻¹ x, tan⁻¹(−x) = − tan⁻¹ x, csc⁻¹(−x) = − csc⁻¹ x.
cos⁻¹(−x) = π − cos⁻¹ x, cot⁻¹(−x) = π − cot⁻¹ x, sec⁻¹(−x) = π − sec⁻¹ x.
Conversion between inverses
Useful conversions (valid in principal ranges):
tan⁻¹ x = sin⁻¹( x / √(1 + x²) ) = cos⁻¹( 1 / √(1 + x²) ) ,
sin⁻¹ x = tan⁻¹( x / √(1 − x²) ) (care with sign when x < 0),
and relations like csc⁻¹ x = sin⁻¹(1/x) (for |x| ≥ 1), sec⁻¹ x = cos⁻¹(1/x) (for |x| ≥ 1).
Addition / subtraction formulas
Arcsine addition (for x,y ∈ [−1,1] and valid principal-value conditions):
sin⁻¹ x + sin⁻¹ y = sin⁻¹( x√(1 − y²) + y√(1 − x²) ),
(similar formula with minus sign for difference).
Arctan addition (most used):
tan⁻¹ x + tan⁻¹ y = tan⁻¹( (x + y) / (1 − xy) ) (if xy < 1; otherwise add or subtract π to get principal value). This formula is very useful for combining angles and for numeric evaluation.
How to think visually / geometrically
Interpret inverse trig functions as angles on the unit circle: if sin θ = x and θ is chosen in the principal range, then θ = sin⁻¹ x. This helps derive conversion formulas: from tan θ = x, you get sin θ = x / √(1 + x²) and cos θ = 1 / √(1 + x²), hence the identities linking arctan with arcsin and arccos.
Use and cautions
Always track domains and principal ranges when simplifying expressions containing inverse trig functions; many apparent algebraic equalities require adjustment by ±π (or not) to remain within the principal value.
- 1) Angle of elevation: If the height of a tower is 50 m and its top subtends an angle θ at a point 100 m away, then θ = tan⁻¹(50/100) = tan⁻¹(0.5). You can convert to sine: θ = sin⁻¹( 0.5 / √(1 + 0.5²) ) = sin⁻¹(0.4472...).
- 2) Combining arctangents: Evaluate tan⁻¹(1/2) + tan⁻¹(1/3). Use formula: (1/2 + 1/3) / (1 − (1/2)(1/3)) = (5/6) / (5/6) = 1, so sum = tan⁻¹(1) = π/4 (no ±π adjustment needed).
- 3) Finding angle from sine reading: A sensor measures the ratio sin θ = 0.6. The principal angle is θ = sin⁻¹(0.6) ≈ 0.6435 rad (≈36.87°). If an application expects an angle outside [−π/2, π/2], you must use unit-circle reasoning to find the correct θ.
- \[Principal ranges: sin⁻¹ x ∈ [−π/2, π/2]\]\[cos⁻¹ x ∈ [0, π]\]\[tan⁻¹ x ∈ (−π/2, π/2)\]\[cot⁻¹ x ∈ (0, π).\]
- \[sin(sin⁻¹ x) = x (−1 ≤ x ≤ 1)\]\[sin⁻¹(sin θ) = θ if θ ∈ [−π/2, π/2].\]
- \[sin⁻¹ x + cos⁻¹ x = π/2.\]
- \[tan⁻¹ x + cot⁻¹ x = π/2.\]
- \[sec⁻¹ x + csc⁻¹ x = π/2 (|x| ≥ 1).\]
- \[tan⁻¹ x = sin⁻¹( x / √(1 + x²) ) = cos⁻¹( 1 / √(1 + x²) ).\]
Algebraic transformations and composition
Algebraic transformations and composition
Core Principle: sin(arcsin x) = x, for x ∈ [−1,1].
Overview
Algebraic transformations are operations on a function f(x) that shift, stretch/compress or reflect its graph. For inverse trigonometric functions (arcsin, arccos, arctan, etc.) these transformations change domain and range in predictable ways. Composition concerns combining functions: f(g(x)). For inverse trig functions two useful kinds appear:
- f(f^{-1}(x)) and f^{-1}(f(x)) — identity-type relations that hold on appropriate domains/ranges (principal values).
- f(g(x)) where g(x) itself is a trig or inverse-trig expression (e.g. arcsin(sin x), arctan(tan x)), often producing piecewise or "sawtooth" profiles because of principal-value restrictions.
Principal branches (important for composition)
- y = arcsin x: domain x ∈ [−1,1], range y ∈ [−π/2, π/2].
- y = arccos x: domain x ∈ [−1,1], range y ∈ [0, π].
- y = arctan x: domain x ∈ ℝ, range y ∈ (−π/2, π/2) (open endpoints).
Algebraic transformations (general rule)
Given y = f(x), the transformed y = A f(Bx + C) + D acts as:
- Horizontal shift: x -> x + h is f(x + h) (shift left by h if h>0). In the argument write Bx + C: horizontal shift = −C/B.
- Horizontal scaling: x -> Bx compresses by factor 1/|B| (B>1 compresses).
- Vertical scaling: multiply by A (stretch if |A|>1, reflect about x-axis if A<0).
- Vertical shift: add D (moves graph up by D).
For inverse trig functions, domain constraints come from the argument. Example: y = arcsin(Bx + C) requires −1 ≤ Bx + C ≤ 1, so x ∈ [(−1−C)/B, (1−C)/B] if B>0 (reverse if B<0). Range is A·[range(f)] + D.
Composition facts
- For x in the domain of f^{-1}, f(f^{-1}(x)) = x. Example: sin(arcsin x) = x for x ∈ [−1,1].
- For x in the range (principal value set) of f, f^{-1}(f(x)) = x. Example: arcsin(sin x) = x only if x ∈ [−π/2, π/2]. Outside that interval the value is the principal-value equivalent (see piecewise rule below).
- Common identity: arcsin x + arccos x = π/2 for x ∈ [−1,1]. Also sin(arcsin x) = x, cos(arccos x) = x, tan(arctan x) = x (on respective domains).
Piecewise / general formula for arcsin(sin x) and similar compositions
Because arcsin returns a value in [−π/2,π/2], arcsin(sin x) reduces x to its principal-value representative. A compact piecewise/algorithmic description:
- arcsin(sin x) = x, if x ∈ [−π/2, π/2].
- More generally, for any integer n, on the interval x ∈ [(2n−1)π/2, (2n+1)π/2] one can write arcsin(sin x) = (−1)^n (x − nπ). (This maps x into [−π/2, π/2].)
- Similarly arctan(tan x) = x for x ∈ (−π/2, π/2), and outside that interval it equals the unique value in (−π/2, π/2) whose tangent equals tan x (producing a periodic sawtooth-like graph with period π).
- arccos(cos x) = x for x ∈ [0, π]; otherwise it is the principal value in [0, π] whose cosine equals cos x (a piecewise reflection every π interval).
Effect on domain and range — quick summary
- For y = A·arcsin(Bx + C) + D: domain: solve −1 ≤ Bx + C ≤ 1; range: y ∈ A·[−π/2, π/2] + D.
- For y = A·arctan(Bx + C) + D: domain: all real x; range: A·(−π/2, π/2) + D (open endpoints).
- For y = A·arccos(Bx + C) + D: domain: −1 ≤ Bx + C ≤ 1; range: A·[0, π] + D.
Practical remark
When composing trig and inverse-trig functions always check whether the inner value lies within the principal domain/range required. If not, express the result in the principal-value form (often piecewise).
- 1) Horizontal shift and vertical scale: y = 2·arcsin(0.5x − 1) + π/6. Domain from −1 ≤ 0.5x − 1 ≤ 1 ⇒ 0 ≤ 0.5x ≤ 2 ⇒ x ∈ [0,4]. Range = 2·[−π/2, π/2] + π/6 = [−π + π/6, π + π/6] = [−5π/6, 7π/6].
- 2) Composition arcsin(sin 5π/6): sin(5π/6) = 1/2, arcsin(1/2) = π/6, so arcsin(sin 5π/6) = π/6 (because 5π/6 is outside [−π/2,π/2], it is mapped to its principal value).
- 3) arctan(tan 3π/4): tan(3π/4) = −1, arctan(−1) = −π/4, so arctan(tan 3π/4) = −π/4 (principal value in (−π/2,π/2)).
- 4) Identity use: For x ∈ [−1,1], arcsin x + arccos x = π/2. Example x = 0.6 → arcsin(0.6) + arccos(0.6) = π/2.
- \[sin(arcsin x) = x\]\[for x ∈ [−1,1].\]
- \[cos(arccos x) = x\]\[for x ∈ [−1,1].\]
- \[tan(arctan x) = x\]\[for x ∈ ℝ.\]
- \[arcsin(sin x) = x\]\[if x ∈ [−π/2, π/2]\]\[more generally arcsin(sin x) = (−1)^n (x − nπ) for x ∈ [(2n−1)π/2\]\[(2n+1)π/2]\]\[n ∈ ℤ.\]
- \[arctan(tan x) = x\]\[if x ∈ (−π/2, π/2) (otherwise equals principal value in (−π/2, π/2)).\]
- \[arcsin x + arccos x = π/2\]\[for x ∈ [−1,1].\]
Differentiation of inverse trigonometric functions
Differentiation of inverse trigonometric functions
Core Principle: d/dx[arcsin x] = 1 / sqrt(1 - x^2), for |x| < 1
Overview: Inverse trigonometric functions (arcsin, arccos, arctan, arccot, arcsec, arccosec) give an angle whose trigonometric value is known. Differentiation of these functions is done using implicit differentiation and the chain rule. Remember each inverse trig function has a principal branch (range) on which it is single-valued and differentiable.
Method (implicit differentiation): If y = f^{-1}(x) then f(y) = x. Differentiate both sides w.r.t x: f'(y)·dy/dx = 1, so dy/dx = 1 / f'(y). For inverse trig functions we often rewrite the relation (for example, y = arcsin x gives sin y = x) and use trigonometric identities to express f'(y) in terms of x.
Key points:
- Use the chain rule for compositions: d/dx[arcsin(u(x))] = u'(x)/sqrt(1 - u(x)^2) (provided |u(x)| < 1).
- Be mindful of domains and ranges: derivatives exist only where the inside argument lies in the domain and the denominator is nonzero.
- Some inverse trig derivatives include absolute values (arcsec, arccosec) because of sign issues coming from the original function's derivative.
Typical derivations (short):
- y = arcsin x ⇒ sin y = x ⇒ cos y · dy/dx = 1 ⇒ dy/dx = 1 / cos y = 1 / sqrt(1 - x^2) for |x| < 1.
- y = arctan x ⇒ tan y = x ⇒ sec^2 y · dy/dx = 1 ⇒ dy/dx = 1 / sec^2 y = 1 / (1 + x^2).
When using results: Always specify the domain of x for which the formula is valid and apply the chain rule if the argument is a function of x.
- Differentiate y = arcsin(2x) for |2x| < 1. Solution: y' = (2) / sqrt(1 - (2x)^2) = 2 / sqrt(1 - 4x^2).
- Differentiate y = arctan(x^2). Solution: y' = (2x) / (1 + x^4) by chain rule since d/dx[arctan u] = u'/(1+u^2) with u = x^2.
- Radar example (real-life): A radar at origin tracks an airplane at coordinates (x(t), h) where h is fixed altitude. Bearing angle θ(t) = arctan(h/x(t)). Then dθ/dt = d/dt[arctan(h/x)] = (1/(1+(h/x)^2))·d/dt(h/x) = (1/(1+h^2/x^2))·(-h x'(t)/x^2) = -h x'(t) / (x^2 + h^2). This gives rate of change of bearing.
- Implicit example (derivation): Let y = arccos x. Then cos y = x ⇒ -sin y · dy/dx = 1 ⇒ dy/dx = -1 / sin y = -1 / sqrt(1 - x^2) for |x| < 1.
- \[d/dx[arcsin x] = 1 / sqrt(1 - x^2)\]\[for |x| < 1\]
- \[d/dx[arccos x] = -1 / sqrt(1 - x^2)\]\[for |x| < 1\]
- \[d/dx[arctan x] = 1 / (1 + x^2)\]\[for all real x\]
- \[d/dx[arccot x] = -1 / (1 + x^2)\]\[for all real x\]
- \[d/dx[arcsec x] = 1 / (|x| sqrt(x^2 - 1))\]\[for |x| > 1\]
- \[d/dx[arccosec x] = -1 / (|x| sqrt(x^2 - 1))\]\[for |x| > 1\]
Integration involving inverse trigonometric functions
Integration involving inverse trigonometric functions
Core Principle: d/dx[arcsin x] = 1 / sqrt(1 - x^2) => ∫ dx / sqrt(1 - x^2) = arcsin x + C
Overview
Integration involving inverse trigonometric functions is the set of techniques and standard results used to evaluate integrals whose antiderivatives are inverse trig functions (arcsin, arctan, arccos, etc.) or integrals that simplify to those standard forms by a suitable substitution. Two common patterns appear:
- Integrands of the form 1/sqrt(a^2 - x^2), 1/(a^2 + x^2), and 1/(x sqrt(x^2 - a^2)), whose antiderivatives are inverse trig (or inverse sec) functions.
- Integrals that contain an inverse trig function as the integrand (for example ∫ arcsin x dx), usually handled by integration by parts.
Key ideas / methods
- Use standard derivatives: the derivative formulas of inverse trig functions give immediate antiderivatives for many integrands.
- Trigonometric substitution: transform algebraic expressions under radicals into trig expressions with substitutions x = a sin θ, x = a tan θ, x = a sec θ to reduce the integrand to a simple trig form.
- Integration by parts: when the integrand is an inverse trig function (e.g. ∫ arctan x dx), choose u = inverse trig and dv = dx.
Common substitutions
- x = a sin θ transforms sqrt(a^2 - x^2) → a cos θ (use when integrand has sqrt(a^2 - x^2)).
- x = a tan θ transforms sqrt(a^2 + x^2) → a sec θ (use when integrand has sqrt(a^2 + x^2) or x^2 + a^2).
- x = a sec θ transforms sqrt(x^2 - a^2) → a tan θ (use when integrand has sqrt(x^2 - a^2)).
Use in problems
These integrals appear while finding areas, arc lengths, centers of mass, potentials and fields in physics, and in geometry (areas of circular segments). Knowing the standard antiderivatives and substitution patterns makes many integrals straightforward.
- Example 1: ∫ dx / sqrt(a^2 - x^2) Solution: Let x = a sin θ, dx = a cos θ dθ, sqrt(a^2 - x^2) = a cos θ. Integral becomes ∫ (a cos θ dθ) / (a cos θ) = ∫ dθ = θ + C. Back-substitute θ = arcsin(x/a). So ∫ dx / sqrt(a^2 - x^2) = arcsin(x/a) + C.
- Example 2: ∫ dx / (a^2 + x^2) Solution: Set x = a tan θ or use standard result. ∫ dx / (a^2 + x^2) = (1/a) arctan(x/a) + C. (Reason: derivative of arctan(x/a) is (1/a) * 1/(1+(x/a)^2) = a/(a^2 + x^2) * 1/a = 1/(a^2 + x^2).)
- Example 3: ∫ arcsin x dx Solution (integration by parts): Let u = arcsin x, dv = dx. Then du = dx / sqrt(1 - x^2), v = x. So ∫ arcsin x dx = x arcsin x - ∫ x * (dx / sqrt(1 - x^2)). Put t = 1 - x^2 => dt = -2x dx, so ∫ x / sqrt(1 - x^2) dx = -sqrt(1 - x^2). Hence ∫ arcsin x dx = x arcsin x + sqrt(1 - x^2) + C.
- Example 4: ∫ arctan x dx Solution (integration by parts): u = arctan x, dv = dx ⇒ du = dx/(1+x^2), v = x. So ∫ arctan x dx = x arctan x - ∫ x/(1+x^2) dx. Let w = 1 + x^2 ⇒ dw = 2x dx. Then ∫ x/(1+x^2) dx = (1/2) ln(1+x^2). Therefore ∫ arctan x dx = x arctan x - (1/2) ln(1 + x^2) + C.
- Example 5: ∫ dx / (x sqrt(x^2 - a^2)) Solution: Put x = a sec θ, dx = a sec θ tan θ dθ, sqrt(x^2 - a^2) = a tan θ. Then integrand = (a sec θ tan θ dθ) / (a sec θ * a tan θ)? Careful simplification gives (1/a) dθ. So integral = (1/a) θ + C. Back-substitute θ = arcsec(x/a) (or θ = arccos(a/x)), so ∫ dx / (x sqrt(x^2 - a^2)) = (1/a) arcsec(|x|/a) + C.
- \[d/dx[arcsin x] = 1 / sqrt(1 - x^2) => ∫ dx / sqrt(1 - x^2) = arcsin x + C\]
- \[d/dx[arccos x] = -1 / sqrt(1 - x^2) => ∫ dx / sqrt(1 - x^2) = -arccos x + C\]
- \[d/dx[arctan x] = 1 / (1 + x^2) => ∫ dx / (1 + x^2) = arctan x + C\]
- \[∫ dx / sqrt(a^2 - x^2) = arcsin(x/a) + C\]
- \[∫ dx / (a^2 + x^2) = (1/a) arctan(x/a) + C\]
- \[∫ dx / sqrt(x^2 - a^2) = ln|x + sqrt(x^2 - a^2)| + C (alternative non-trig antiderivative)\]
Solving equations and inequalities
Solving equations and inequalities
Core Principle: Definitions (principal values): arcsin: [-1,1] → [-π/2, π/2]; arccos: [-1,1] → [0, π]; arctan: ℝ → (-π/2, π/2).
What this topic covers
Solving equations and inequalities that involve inverse trigonometric functions (arcsin, arccos, arctan, arccot, arcsec, arccosec). The main ideas are: convert the inverse-trig statement to a trig statement using the definitions, respect principal-value ranges, use monotonicity to transform inequalities, and apply algebraic/trigonometric identities (including addition formulas for arctan) with attention to branch/period adjustments.
Principal-value ranges (important)
- y = arcsin x ⇒ y ∈ [-π/2, π/2], x ∈ [-1,1]
- y = arccos x ⇒ y ∈ [0, π], x ∈ [-1,1]
- y = arctan x ⇒ y ∈ (-π/2, π/2), x ∈ ℝ
General approach to equations
1) If you have y = arcsin(f(x)) = A, rewrite as f(x) = sin A and ensure A lies in the principal range of arcsin. If A does not lie in the principal range, either restrict or use the general solution of sin θ = value with period/alternate signs.
2) For equations that sum inverse functions (e.g., arctan u + arctan v = some value), use the appropriate addition formula for inverse tangents with the extra ±π adjustment where required.
3) Always check (a) domain restrictions (e.g., argument of arcsin/arccos must be in [-1,1]) and (b) whether any branch/period adjustment (±π, or sign flip for sin) is needed when converting between inverse and direct trig forms.
Solving inequalities
- Use monotonicity: arcsin x and arctan x are strictly increasing on their domains, and arccos x is strictly decreasing on [-1,1]. Thus
- arcsin x > a ⇔ x > sin a provided a ∈ [-π/2,π/2]
- arccos x < a ⇔ x > cos a provided a ∈ [0,π]
Common pitfalls
- Forgetting domain of inverse trig arguments (e.g., arcsin argument must be in [-1,1]).
- Applying sin/cos/tan to both sides without checking that both sides lie in a region where the trig function is one-to-one (or else adding periodic solutions).
- Not adding the necessary ±π (or nπ) corrections when turning an inverse-trig sum into a single expression using addition formulas.
Worked method summary
Step 1: Identify principal ranges and domain of expressions.
Step 2: Convert inverse-trig equality to a trig equality: e.g., arcsin(g(x)) = A ⇒ g(x) = sin A and require A ∈ [-π/2,π/2].
Step 3: Solve the algebraic equation from Step 2 and verify domain and branch conditions.
Step 4: For inequalities, use monotonicity or reduce to trig inequalities with care about periodic solutions.
- 1) Solve arcsin(2x) = π/6. Solution: 2x = sin(π/6) = 1/2 ⇒ x = 1/4. Check domain: |2x| ≤ 1 ⇒ |1/2| ≤ 1 OK. So x = 1/4.
- 2) Solve arcsin x + arccos x = π/2. For all x ∈ [-1,1] we have arcsin x + arccos x = π/2 (identity). Therefore every x in [-1,1] is a solution.
- 3) Solve inequality arcsin x > 0. Since arcsin is strictly increasing on [-1,1] and arcsin 0 = 0, the inequality is equivalent to x > 0 with x ∈ [-1,1]. So solution: 0 < x ≤ 1.
- 4) Solve arctan x + arctan 2x = π/4. Using arctan addition formula: arctan u + arctan v = arctan((u+v)/(1-uv)) + kπ (choose k so result in principal range). Here (u+v)/(1-uv) = (3x)/(1-2x^2). Taking tangent on both sides gives (3x)/(1-2x^2) = 1 ⇒ 2x^2 + 3x - 1 = 0 ⇒ x = (-3 ± √17)/4. Each root must be checked to ensure the arctan-sum gives π/4 (no extra ±π).
- \[Definitions (principal values): arcsin: [-1,1] → [-π/2, π/2]\]\[arccos: [-1,1] → [0, π]\]\[arctan: ℝ → (-π/2, π/2).\]
- \[sin(arcsin x) = x for x ∈ [-1,1]\]\[cos(arccos x) = x for x ∈ [-1,1]\]\[tan(arctan x) = x for x ∈ ℝ.\]
- \[Identity: arcsin x + arccos x = π/2 for x ∈ [-1,1].\]
- \[arctan addition formula: arctan u + arctan v = arctan((u+v)/(1-uv)) + kπ\]\[where k ∈ {0, ±1} chosen so the RHS lies in (-π/2, π/2) (or matches the required branch).\]
- \[General solutions for trig equations (use when converting inverse-trig to trig): sin θ = a ⇒ θ = nπ + (-1)^n α\]\[where α = arcsin a (principal) and n ∈ ℤ\]\[cos θ = a ⇒ θ = 2nπ ± arccos a\]\[tan θ = a ⇒ θ = arctan a + nπ.\]
- \[Monotonicity rules for inequalities: arcsin and arctan are strictly increasing on their domains\]\[arccos is strictly decreasing on [-1,1].\]
Continuity and monotonicity
Continuity and monotonicity
Core Principle: arcsin x: domain = [−1,1], range = [−π/2, π/2]; d/dx(arcsin x) = 1/√(1−x²), |x|<1
Definition (continuity): A function f is continuous at x0 if lim_{x→x0} f(x) = f(x0). For inverse trigonometric functions we check continuity on their principal domains (where they are single-valued).
Definition (monotonicity): A function is monotone increasing on an interval if x1 Principal inverse trig functions — continuity and monotonicity: Why monotone implies inverse exists (on restricted intervals): A continuous strictly monotone function on an interval has a continuous inverse on the image of that interval. The standard inverse trig functions come from restricting sin, cos, tan, cot to intervals where they are one-to-one and continuous. Notes about endpoints and derivatives: Continuity at endpoints (for arcsin/arccos) holds even though the derivative becomes unbounded there. So continuity is about limits, not finiteness of derivative.
- arccos(1/2) = π/3. Since arccos is decreasing, values >1/2 give angles <π/3 and values <1/2 give angles >π/3.
- arcsin(−√2/2) = −π/4. arcsin is increasing: arcsin(−1) = −π/2 < −π/4 < arcsin(0) = 0.
- arctan(0.75) ≈ 0.6435 rad (≈36.87°). Use when converting a slope (rise/run = 0.75) to angle of inclination: θ = arctan(0.75).
- If vectors u and v have dot product u·v = |u||v|·(1/2), angle between them = arccos(1/2) = π/3 (used in engineering/design).
- \[arcsin x: domain = [−1,1]\]\[range = [−π/2, π/2]\]\[d/dx(arcsin x) = 1/√(1−x²), |x|<1\]
- \[arccos x: domain = [−1,1]\]\[range = [0, π]\]\[d/dx(arccos x) = −1/√(1−x²), |x|<1\]
- \[arctan x: domain = R\]\[range = (−π/2, π/2)\]\[d/dx(arctan x) = 1/(1+x²)\]
- \[arccot x: domain = R\]\[range = (0, π) (principal)\]\[d/dx(arccot x) = −1/(1+x²)\]
- \[arcsec x: domain = (−∞,−1] ∪ [1,∞)\]\[d/dx(arcsec x) = 1/(|x| √(x²−1))\]\[for |x|>1\]
- \[arccsc x: domain = (−∞,−1] ∪ [1,∞)\]\[d/dx(arccsc x) = −1/(|x| √(x²−1))\]\[for |x|>1\]
Principal-value issues and branch behavior
Principal-value issues and branch behavior
Core Principle: Principal ranges: arcsin x ∈ [−π/2, π/2], arccos x ∈ [0, π], arctan x ∈ (−π/2, π/2).
What is the issue? The trigonometric functions (sin, cos, tan ...) are periodic and not one-to-one on R, so their inverses are multivalued: if y = sin x then x = arcsin(y) + 2πn or x = π - arcsin(y) + 2πn. For a well-defined inverse we pick a single “branch” (a conventionally chosen interval) called the principal branch, and its values are the principal values.
Principal ranges (conventions used in Class 12):
- arcsin x (principal value): domain x ∈ [−1,1], range y ∈ [−π/2, π/2]
- arccos x: domain x ∈ [−1,1], range y ∈ [0, π]
- arctan x: domain x ∈ R, range y ∈ (−π/2, π/2)
- arccot x (common convention): range y ∈ (0, π) (some texts use (−π/2, π/2) — be consistent)
Branch behavior and discontinuities: The principal branch is chosen so the inverse is single-valued and continuous on its domain, but different principal branches for different inverse functions have different end points. For example arctan x has horizontal asymptotes at ±π/2 and is continuous on R, whereas arcsin x and arccos x are defined only on [−1,1]. Because trig functions are periodic the full set of solutions to an equation includes adding integer multiples of the period (often π or 2π); the principal value is just the one value from the chosen range.
Practical consequence: When solving trig equations you must (a) give the principal value if asked, and (b) give general solutions by adding the appropriate period or using symmetry (e.g. sin x = a has two solutions in [0,2π)). When using inverse trig identities (like addition formulas for arctan) you must check and sometimes add or subtract π to move the computed principal value into the correct quadrant — this is the common "branch correction".
Why it matters in real life: In engineering and physics we often need the actual physical angle (not just the principal value) — e.g. compass headings, robot joint angles or signal phase. Software that returns an inverse-trig principal value (like atan2) includes rules to choose the correct branch/quadrant to avoid ambiguity.
- Example 1 — arcsin: sin x = 1/2. Principal value: arcsin(1/2) = π/6. General solutions: x = π/6 + 2πn or x = 5π/6 + 2πn (n ∈ Z). The principal value picks π/6 (in [−π/2, π/2]).
- Example 2 — arctan and periodicity: tan x = 1. Principal value: arctan(1) = π/4. General solutions: x = π/4 + nπ (n ∈ Z). The function arctan returns only the principal one in (−π/2, π/2).
- Example 3 — arctan sum/branch correction: tan^{-1}2 + tan^{-1}3 using formula tan^{-1}x + tan^{-1}y = tan^{-1}((x+y)/(1−xy)) requires care because 1−xy = 1−6 = −5 < 0; the right-hand expression gives a principal value that must be adjusted by +π to get the true sum (which is 3π/4).
- Example 4 — applied: In polar conversion from (x,y) to (r,θ) we compute θ = arctan(y/x) but must check signs of x and y to place θ in the correct quadrant; computer languages provide atan2(y,x) to handle the branch correctly.
- \[Principal ranges: arcsin x ∈ [−π/2, π/2]\]\[arccos x ∈ [0, π]\]\[arctan x ∈ (−π/2, π/2).\]
- \[General solutions: If y = arcsin(a) then solutions for x in sin x = a are x = arcsin(a) + 2πn or x = π − arcsin(a) + 2πn\]\[for cos x = a: x = ±arccos(a) + 2πn\]\[for tan x = a: x = arctan(a) + nπ.\]
- \[Derivatives (on principal branch interior): d/dx[arcsin x] = 1/√(1−x^2)\]\[d/dx[arccos x] = −1/√(1−x^2)\]\[d/dx[arctan x] = 1/(1+x^2).\]
- \[Arctan addition (watch branch): tan^{-1}x + tan^{-1}y = tan^{-1}((x+y)/(1−xy)) but if 1−xy < 0 you must add or subtract π to place the result in the correct principal range\]\[similarly handle special quadrant cases.\]
- \[Identity inside domain: arcsin x + arccos x = π/2 (for x ∈ [−1,1]).\]
Applications and typical problem types
Applications and typical problem types
Core Principle: Principal ranges: arcsin x ∈ [−π/2, π/2], arccos x ∈ [0, π], arctan x ∈ (−π/2, π/2).
Inverse trigonometric functions (arcsin, arccos, arctan, etc.) map a ratio back to an angle. Their study in Class 12 focuses on algebraic manipulation, solving equations/inequalities, calculus applications (differentiation/integration), and modelling simple geometrical/real-life situations by recovering angles from measured ratios. Key ideas are principal values (single‑valued branches), triangle/substitution methods (to convert trig of an inverse‑trig into algebraic expressions), and addition/combination formulae (especially for arctan).
Typical problem types fall into these categories:
- Simplification/identity verification using triangle representation (e.g., express sin(arctan x) in algebraic form).
- Solving equations involving sums/differences of inverse trig functions (use addition formulae or tangent addition and check branch conditions).
- Finding principal values and solving inequalities (use monotonicity and ranges).
- Calculus: differentiate and integrate functions containing inverse trig (use standard derivative/integral forms and chain rule/substitution).
- Limits involving inverse trig (use series or small‑angle approximations, or convert to algebraic forms).
- Applications: geometry/triangles, heights & distances, slopes and angles of elevation/inclination, robotics/joint angles, navigation bearings — find angle from measured ratio using inverse trig.
Solution strategies (practical rules):
- Always note principal ranges: arcsin ∈ [−π/2, π/2], arccos ∈ [0, π], arctan ∈ (−π/2, π/2). Use them to check/choose correct branches.
- Use right‑triangle substitution: if θ = arctan x, draw triangle with opposite = x, adjacent = 1 so hypotenuse = √(1+x^2); then compute sinθ, cosθ etc. This converts trig of inverse trig to algebraic expressions.
- For sums like arctan a + arctan b, use arctan addition formula but check the sign/π adjustments depending on quadrants (product conditions). For equations, verify roots satisfy domain/branch restrictions.
- For differentiation/integration, memorize standard derivatives and integrals of inverse trig and apply chain rule/substitution (e.g., d/dx[arcsin u] = u'/(√(1−u^2))).
- 1) Simplify sin(arctan x). Solution idea: Let θ = arctan x ⇒ tanθ = x = opposite/adjacent. Take opposite = x, adjacent = 1, hypotenuse = √(1+x^2). So sinθ = opposite/hypotenuse = x/√(1+x^2).
- 2) Solve arctan x + arctan(2x) = π/4. Use arctan addition formula: arctan((3x)/(1−2x^2)) = π/4 ⇒ (3x)/(1−2x^2) = 1 (ensure branch condition 1−2x^2>0). Solve 2x^2+3x−1=0 ⇒ x = (−3 ± √17)/4. Check branch: only x = (−3+√17)/4 ≈ 0.281 is valid.
- 3) Differentiate y = arcsin(x^2). y' = (2x)/√(1−x^4) (chain rule: d/dx[arcsin u] = u'/(√(1−u^2))).
- 4) Integrate ∫ dx/(1+x^2). Result: arctan x + C. (Standard integral often used to transform rational expressions to inverse trig.)
- 5) Limit: lim_{x→0} (arcsin x)/x = 1, since arcsin x ~ x for small x (use series or definition).
- 6) Application (height & distance): If a tower of height h seen from point P makes angle of elevation α, and horizontal distance is d, then tan α = h/d so α = arctan(h/d). Given h and d you compute α using arctan.
- \[Principal ranges: arcsin x ∈ [−π/2, π/2]\]\[arccos x ∈ [0, π]\]\[arctan x ∈ (−π/2, π/2).\]
- \[Basic compositions: sin(arcsin x)=x\]\[cos(arccos x)=x\]\[tan(arctan x)=x (within domains).\]
- \[Triangle conversions: If θ = arctan x ⇒ sinθ = x/√(1+x^2)\]\[cosθ = 1/√(1+x^2)\]\[secθ = √(1+x^2)\]\[If θ = arcsin x ⇒ tanθ = x/√(1−x^2)\]\[cosθ = √(1−x^2).\]
- \[Arctan addition: arctan a + arctan b = arctan((a+b)/(1−ab)) (use with care: add π or −π when necessary depending on signs/branches).\]
- \[Special identity: arctan x + arctan(1/x) = π/2 for x>0, = −π/2 for x<0 (x≠0).\]
- \[Derivatives: d/dx[arcsin x] = 1/√(1−x^2)\]\[d/dx[arccos x] = −1/√(1−x^2)\]\[d/dx[arctan x] = 1/(1+x^2).\]
Useful formulas and summary table
Useful formulas and summary table
Core Principle: Domain & Range: arcsin x: x∈[−1,1], range [−π/2,π/2]; arccos x: x∈[−1,1], range [0,π]; arctan x: x∈R, range (−π/2,π/2).
This summary collects the essential definitions, principal value ranges, algebraic relations, identities, differentiation & integration rules and useful transformation formulas for inverse trigonometric functions (Class 12 CBSE). Inverse trigonometric functions give angles from given trigonometric ratios and are defined with principal value (PV) ranges to make them single-valued.
- Principal values and domains
- y = arcsin x (or sin^{-1}x): domain x ∈ [−1,1], range y ∈ [−π/2, π/2]
- y = arccos x (or cos^{-1}x): domain x ∈ [−1,1], range y ∈ [0, π]
- y = arctan x (or tan^{-1}x): domain x ∈ (−∞, ∞), range y ∈ (−π/2, π/2)
- y = arccot x: domain x ∈ (−∞, ∞), range y ∈ (0, π)
- y = arcsec x: domain |x| ≥ 1, range y ∈ [0, π] \ {π/2}
- y = arccsc x: domain |x| ≥ 1, range y ∈ [−π/2, π/2] \ {0}
- Basic compositions & symmetry
- sin(arcsin x) = x for x ∈ [−1,1]
- arcsin(sin x) = x for x ∈ [−π/2, π/2] (equals the PV of the angle otherwise)
- Similar: cos(arccos x)=x, arccos(cos x)=x for x ∈ [0,π]; tan(arctan x)=x, arctan(tan x)=x for x ∈ (−π/2,π/2)
- Odd/even properties: arcsin(−x)=−arcsin x, arctan(−x)=−arctan x, arccos(−x)=π−arccos x
- Key identities
- arcsin x + arccos x = π/2 for x ∈ [−1,1]
- arctan x + arccot x = π/2 for all real x
- arcsec x + arccsc x = π/2 (with principal value conventions)
- Addition / subtraction (useful algebraic forms)
- arcsin u ± arcsin v = arcsin( u√(1−v^2) ± v√(1−u^2) ) (valid when the RHS value lies in PV range; otherwise adjust by ±π)
- arccos u ± arccos v = arccos( uv ∓ √((1−u^2)(1−v^2)) ) (check range)
- arctan u + arctan v = arctan((u+v)/(1−uv)) + kπ, where k = 0 or ±π chosen so result lies in (−π/2, π/2)
- 2·arctan x = arctan(2x/(1−x^2)) with range corrections when denominator ≤ 0
- Derivatives & integrals (standard)
- d/dx[arcsin x] = 1/√(1−x^2), x ∈ (−1,1)
- d/dx[arccos x] = −1/√(1−x^2)
- d/dx[arctan x] = 1/(1+x^2)
- d/dx[arccot x] = −1/(1+x^2)
- d/dx[arcsec x] = 1/(|x|√(x^2−1)), d/dx[arccsc x] = −1/(|x|√(x^2−1))
- ∫ dx/√(1−x^2) = arcsin x + C, ∫ dx/(1+x^2) = arctan x + C
Use this table as a quick reference. Beware of branch (principal value) adjustments when combining inverse functions — many algebraic formulas require checking which multiple of π to add/subtract so the result lies in the principal range.
- Surveying/height measurement: If a surveyor measures the angle of elevation θ of the top of a tower from a point at horizontal distance d and the tower height is h, then θ = arctan(h/d). Rearranged: h = d·tanθ.
- GPS / satellite elevation: Elevation angle α from horizontal given height h and slant distance r: α = arcsin(h/r).
- Robotics / inverse kinematics: To point an end-effector toward coordinates (x,y) in the plane, a joint angle can be computed as θ = arctan2(y,x) (a two-argument arctan variant that uses quadrants).
- Phase difference in signals: If tan φ = X/Y (ratio of sine and cosine components), the phase angle φ is φ = arctan(X/Y) (choose correct quadrant).
- Solving equations: Solve sin θ = 0.6 → θ = arcsin(0.6) ≈ 0.6435 rad (principal value). All solutions: θ = arcsin(0.6) + 2kπ or θ = π − arcsin(0.6) + 2kπ.
- \[Domain & Range: arcsin x: x∈[−1,1]\]\[range [−π/2,π/2]\]\[arccos x: x∈[−1,1]\]\[range [0,π]\]\[arctan x: x∈R\]\[range (−π/2,π/2).\]
- \[Composition: sin(arcsin x)=x (x∈[−1,1])\]\[arcsin(sin x)=x for x∈[−π/2,π/2].\]
- \[Symmetry: arcsin(−x)=−arcsin x\]\[arctan(−x)=−arctan x\]\[arccos(−x)=π−arccos x.\]
- \[Relation: arcsin x + arccos x = π/2.\]
- \[Arctan/arccot: arctan x + arccot x = π/2.\]
- \[arctan addition: arctan u + arctan v = arctan((u+v)/(1−uv)) + kπ (choose k so result in PV).\]
Key Concepts
- Inverse trigonometric function
- A function that reverses a trigonometric function on a restricted domain so that each output corresponds to a unique angle.
- Principal value
- The single value chosen from the multi-valued inverse-trigonometric outputs by restricting the range (principal branch).
- Domain
- The set of input values for which an inverse trigonometric function is defined.
- Range
- The set of output (angle) values an inverse trigonometric function can produce (principal range).
- arcsin (sin⁻¹)
- The inverse of sine on the principal range [−π/2, π/2]; returns an angle whose sine is the given number.
- arccos (cos⁻¹)
- The inverse of cosine on the principal range [0, π]; returns an angle whose cosine is the given number.
- arctan (tan⁻¹)
- The inverse of tangent on the principal range (−π/2, π/2); returns an angle whose tangent is the given number.
- arccot (cot⁻¹)
- The inverse of cotangent on a chosen principal range (commonly (0, π)); returns an angle whose cotangent is the given number.
- arcsec (sec⁻¹)
- The inverse of secant; principal range usually [0, π] with π/2 excluded. Returns an angle whose secant equals the given value.
- arccosec / arccsc (cosec⁻¹)
- The inverse of cosecant; principal range often [−π/2, π/2] with 0 excluded. Returns an angle whose cosecant equals the given value.
- Composition identity
- Relationships showing inverse and original functions compose to the identity on appropriate domains.
- Even / Odd properties
- Parity behavior of inverse trig functions: some are odd (f(−x)=−f(x)), others follow specific symmetry rules.
- Monotonicity
- Inverse trigonometric functions on their principal ranges are monotonic (either strictly increasing or decreasing).
- Continuity
- Inverse trigonometric functions are continuous on their principal domains (except at points excluded from domain).
- Differentiability
- Inverse trig functions are differentiable on interior points of their domains; derivatives may be undefined or infinite at endpoints.
- Derivative of arcsin
- d/dx [arcsin x] = 1 / √(1 − x²) for x in (−1,1).
- Derivative of arctan
- d/dx [arctan x] = 1 / (1 + x²) for all real x.
- Derivative of arccos
- d/dx [arccos x] = −1 / √(1 − x²) for x in (−1,1).
- Basic integration formulas
- Integrals that yield inverse trig functions, used to invert derivatives or evaluate antiderivatives.
- Arctan addition formula
- A useful identity: for uv < 1, arctan u + arctan v = arctan((u + v)/(1 − uv)), adjusted by ±π when needed.
Practice Questions
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State the principal-value range of arcsin x and arccos x. / arcsin x तथा arccos x की मुख्य-मान परास लिखिए।
Show answer
arcsin x has range [-π/2, π/2] and arccos x has range [0, π], both with domain [-1, 1]. / arcsin x की परास [-π/2, π/2] तथा arccos x की परास [0, π] है, दोनों का प्रांत [-1, 1] है।
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Prove that arcsin x + arccos x = π/2 for x ∈ [-1,1]. / सिद्ध कीजिए कि x ∈ [-1,1] के लिए arcsin x + arccos x = π/2।
Show answer
Let arcsin x = θ, so x = sin θ = cos(π/2 - θ); since π/2-θ ∈ [0,π], arccos x = π/2 - θ, giving the sum π/2. / मान लीजिए arcsin x = θ, तो x = sin θ = cos(π/2 - θ); चूँकि π/2-θ ∈ [0,π], अतः arccos x = π/2 - θ, और योग π/2 प्राप्त होता है।
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Evaluate arcsin(sin 5π/6). / arcsin(sin 5π/6) का मान ज्ञात कीजिए।
Show answer
sin(5π/6) = 1/2 and 5π/6 lies outside [-π/2, π/2], so arcsin(sin 5π/6) = arcsin(1/2) = π/6. / sin(5π/6) = 1/2 तथा 5π/6, [-π/2, π/2] के बाहर है, अतः arcsin(sin 5π/6) = arcsin(1/2) = π/6।
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Evaluate tan^{-1}(1/2) + tan^{-1}(1/3) using the addition formula. / योग सूत्र से tan^{-1}(1/2) + tan^{-1}(1/3) का मान ज्ञात कीजिए।
Show answer
(1/2+1/3)/(1-(1/2)(1/3)) = (5/6)/(5/6) = 1; since xy<1, the sum = tan^{-1}(1) = π/4. / (1/2+1/3)/(1-(1/2)(1/3)) = (5/6)/(5/6) = 1; चूँकि xy<1, योग = tan^{-1}(1) = π/4।
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Why must the domain of sin x be restricted before defining arcsin x? / arcsin x को परिभाषित करने से पहले sin x का प्रांत प्रतिबंधित क्यों किया जाता है?
Show answer
Because sin x is periodic and not one-one on R; restricting it to [-π/2, π/2] makes it one-one so a single-valued inverse exists. / क्योंकि sin x आवर्ती है और R पर एकैकी नहीं; इसे [-π/2, π/2] तक सीमित करने से वह एकैकी बनता है जिससे एकल-मान व्युत्क्रम परिभाषित होता है।
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Differentiate y = arcsin(2x) with respect to x. / x के सापेक्ष y = arcsin(2x) का अवकलन कीजिए।
Show answer
Using d/dx[arcsin u] = u'/√(1-u^2) with u=2x: y' = 2/√(1 - 4x^2), valid for |2x|<1. / d/dx[arcsin u] = u'/√(1-u^2) में u=2x रखने पर: y' = 2/√(1 - 4x^2), जो |2x|<1 के लिए मान्य है।
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Evaluate ∫ dx/(a^2 + x^2). / ∫ dx/(a^2 + x^2) का मान ज्ञात कीजिए।
Show answer
∫ dx/(a^2 + x^2) = (1/a) arctan(x/a) + C, since d/dx[arctan(x/a)] = 1/(a^2+x^2). / ∫ dx/(a^2 + x^2) = (1/a) arctan(x/a) + C, क्योंकि d/dx[arctan(x/a)] = 1/(a^2+x^2)।
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Find arcsec(-2) using its relation to arccos. / arccos से संबंध का उपयोग कर arcsec(-2) ज्ञात कीजिए।
Show answer
arcsec(-2) = arccos(1/(-2)) = arccos(-1/2) = 2π/3, lying in [0,π]\{π/2}. / arcsec(-2) = arccos(1/(-2)) = arccos(-1/2) = 2π/3, जो [0,π]\{π/2} में है।
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